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September 30th, 2018, 10:56 AM  #1 
Newbie Joined: Jan 2018 From: Belgium Posts: 7 Thanks: 0  Test divisibility polynomials
Hello everybody, I am currently revising some math out of pure interest. I found some interesting books, unfortunately not always with solutions to the problems. A problem I am stuck with is the following: "Proof that the polynomial n^3 + (n+1)^3 + (n+2)^2 is divisible by 9". I do not know if there is a general test for this one. My approach would be the try to work out this polynomial as far as possible and try to get a factor 9 out with the distributive property. I always arrive at 3n^3 + 9n^2 + 27n + 9; seeming to prove that it is divisible by 3,but not by 9. It would seem something is off with the coefficient of n^3. Could anyone provide me some insight in this problem of offer another approach please? Thanks in advance! 
September 30th, 2018, 11:54 AM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry 
I take it that the last term is meant to be $(n+2)^3$ rather than $(n+2)^2$? If so, it actually expands out as $3n^3 + 9n^2 + 15n + 9$. Now this is divisible by $9$ if and only if $3n^3 + 15n$ is divisible by $9$. But this is equivalent to $n^3 + 5n = n(n^2 + 5)$ being divisible by $3$. See if you can show this. (Hint: $n$ is either a multiple of $3$, one more than a multiple of $3$, or one less than a multiple of $3$. You might like to deal with each case separately.) 
September 30th, 2018, 04:41 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 621 Thanks: 394 Math Focus: Dynamical systems, analytic function theory, numerics 
Another way to think about it is you polynomial is the sum of 3 consecutive cubes. The cubes (in order from 0 to 8 ) mod 9 are $\{0,1,8,0,1,8,0,1,8\}$.

September 30th, 2018, 05:24 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
Then $n^3 = 3^3 = 27, \ (n + 1)^3 = 4^3 = 64, \text { and } (n + 2)^2 = 5^2 = 25 \implies$ $n^3 + (n + 1)^3 + (n + 2)^2 = 27 + 64 + 25 = 116.$ And 9 does not divide evenly into 116. So the problem as you have given it is invalid. So you cannot prove what you say you are to prove. Actually, to have the problem make any sense, you needed to specify that n is an integer and that the polynomial divided by 9 is an integer because, without those restrictions, 9 divides every number. Here is a reasonable problem that you may have been given: $\text {Given } n \in \mathbb Z \text { and } x = n^3 + (n + 1)^3 + (n + 2)^3,$ $\text {prove that } \dfrac{x}{9} \in \mathbb Z.$ There are several ways to attack such a problem. I would likely notice that either n, n + 1, or n + 2 will necessarily be a multiple of 3 if n is an integer. That gives three cases. $\text {CASE I: } \dfrac{n}{3} = k \in \mathbb Z \implies$ $n = 3k,\ n + 1 = 3k + 1, \text { and } n + 2 = 3k + 2 \implies$ $x = (3k)^3 + (3k + 1)^3 + (3k + 2)^3 =$ $(3k)^3 + (3k)^3 + 3(3k)^2(1) + 3(3k)(1^2) + 1^3 + (3k)^3 + 3(3k)^2(2) + 3(3k)(2^2) + 2^3 =$ $27k^3 + 27k^3 + 27k^2 + 9k + 1 + 27k^3 + 54k^2 + 36k + 8 =$ $81k^3 + 81k^2 + 45k + 9 = 9(9k^3 + 9k^2 + 5k + 1).$ $\therefore \dfrac{x}{9} = 9k^3 + 9k^2 + 5k + 1 \in \mathbb Z \ \because \ k \in \mathbb Z.$ I'll let you do the other two cases. This may not be an elegant solution, but when I see n, n + 1, and n + 2, with n an integer, I know one of those numbers is divisible by 3. That gives me an intuitive line of attack. Cjem and SDK started from somewhat different and eventually simpler lines of attack, but they are much, much more clever than I am. But there are different ways to attack a problem Find the one that works for you. Last edited by JeffM1; September 30th, 2018 at 05:34 PM.  
October 1st, 2018, 04:23 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 
If (3p) + (3q + 1)³ + (3r + 2)³ is expanded, every coefficient of the resulting polynomial is divisible by 9. Hence the result holds for the cubes of any three integers that happen to be congruent to 0, 1 and 2 modulo 3.

October 1st, 2018, 08:00 AM  #6  
Newbie Joined: Jan 2018 From: Belgium Posts: 7 Thanks: 0  Quote:
 

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